Area Between Curves

Subtract the Lower Curve from the Upper

Right end b1

👀 See It

①The area between two curves is the integral of (upper − lower)

②On [0,1], y=x is upper and y=x² is lower

③Growing b to 1 fills the area to 1/6

The Area Formula

Area between two curves

S = ∫_a^b (f(x) − g(x)) dx (f ≥ g)

a, b are the x-coordinates of the intersections; f is the upper, g the lower curve

Find the Intersections First

Integration limits = intersections

f(x) = g(x)

Solving the two equations gives the x-values that become the limits a and b

Compute It Directly

Example 1

Find the area enclosed by y=x and y=x².

Find the intersections — solve the two equations.

x = x² ⇒ x = 0, 1

On [0,1] the line x is upper, so integrate (x − x²).

S = ∫₀¹ (x − x²) dx = [x²2 − x³3]₀¹ = 16

▸ 1/6

Plug one point in the interval to check which curve is upper and avoid sign errors.

Example 2

Find the area enclosed by y=x² and the line y=2x.

Find the intersections.

x² = 2x ⇒ x = 0, 2

On [0,2] the line 2x is upper, so integrate (2x − x²).

S = ∫₀² (2x − x²) dx = [x² − x³3]₀² = 43

▸ 4/3

For a curve and a line, the (upper − lower) principle is exactly the same.

Wrap-up

Core strategy

S = ∫_a^b (upper − lower) dx, a·b are intersections

Set the limits by the intersections, then integrate upper minus lower

2022 provincial mock exam Math type, adapted

What is the area enclosed by y=x²−2x and the x-axis?

①2/3

②1

③4/3

④2

⑤8/3

▸ ③ 4/3

Find the intersections with the x-axis (y=0).

x² − 2x = 0 ⇒ x = 0, 2

On [0,2] the curve is below the axis, so integrate (0 − (x²−2x)).

S = ∫₀² (2x − x²) dx = 43

🎯 Exam Points

①Area = ∫(upper − lower)dx

②The limits must be the intersections

③If the curve is below the axis, integrate (0 − curve) = −curve

④If upper and lower swap, split the interval

⑤You may write it with an absolute value, but check the sign on each piece

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